Question

Cystic fibrosis is a lung disorder that often results in death. It is inherited but can be inherited only if both parents are carriers of an abnormal gene. In 1989, the CF gene that is abnormal in carriers of cystic fibrosis was identified. The probability that a randomly chosen person of European ancestry carries an abnormal CF gene is 1/25. (The probability is less in other ethnic groups.) The CF20m test detects most but not all mutations of the CF gene. The test is positive for 90% of people who are carriers. It is (ignoring human error) never positive for people who are not carriers.

a. Jason tests positive. What is the probability that he is a carrier?

b. What is the conditional probability that anyone tests negative?

Answer 1

Solution Probability that Jason is a carrier given he tests positive

Let P and C respectively represent the events that a person tests positive and the person is a carrier.

Trivially, then P^C and C^C respectively represent the events that a person tests negative and the person is not a carrier…………………………………………………………………………………….. (1)

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)……....….(2)

P(B) = {P(B/A) x P(A)} + {P(B/A^C) x P(A^C)}…………………………………          .(3)

P(A/B) = P(B/A) x {P(A)/P(B)}……………………………..…………………..…….(4)

Now to work out the solution,

Vide (1), given ‘The probability that a randomly chosen person of European ancestry carries an abnormal CF gene is 1/25.’ => P(C) =1/25 = 0.04 …………………………………. (5)

‘The test is positive for 90% of people who are carriers.’ => P(P/C) = 0.90 …………(6)

‘It is never positive for people who are not carriers.’ => P(P/C^C) = 0 ……..…………(7)

(5) => P(C^C) = 1 – 0.04 = 0.96 …………………………………………………………(8)

(6) => P(P^C/C) = 1 – 0.9 = 0.1 …………………………………………………………(9)

(7) => P(P^C/C^C) = 1 – 0 = 1.0 …………………………………………………………(10)

Part (a)

Probability that Jason is a carrier given that he tests positive

= P(C/P)

= P(P/C) x P(C)/P(P) [vide (4)]

= 0.9 x 0.0.04/P(P) [vide (6) and (5)] ………………………………………………………. (11)

Now, vide (3),

P(P) = {P(P/C) x P(C)} + {P(P/C^C) x P(C^C)}

= (0.9 x 0.04) + (0 x 0.96) [vide (6), (5), (7) and (8)]

= 0.036 ………………………………………………………………………………………….(12)

(12) in (11) gives: P(C/P) = 0.036/0.036 = 1.

Thus, Probability that Jason is a carrier given that he tests positive = 1 Answer

Part (b)

Probability that anyone tests negative

= P(P^C)

= {P(P^C/C) x P(C)} + {P(P^C/C^C) x P(C^C)} [vide (3)]

= (0.1 x 0.04) + (1 x 0.96) [vide (9), (5), (10) and (8)]

= 0.964 Answer

DONE